Question

In triangle abc, m of acb = 90, cd is perpendicular to ab , m of acd is 60. and bd is 5 cm. find ad

Answer 1

Let us draw a picture to make the things more clear.

Attached is the image.

We have been given that

$\angle acd = 60 ^{\circ}$

Therefore, we have

$\angle dcb =90- 60= 30 ^{\circ}$

Now, in triangle bcd, we have

$\tan30 = \frac{5}{cd}\\ \\ \frac{1}{\sqrt 3}=\frac{5}{cd}\\ \\ cd=5\sqrt 3$

Now, in triangle acb, we have

$tan 60 = \frac{ad}{5\sqrt3} \\ \\ \sqrt 3= \frac{ad}{5\sqrt3}\\ \\ ad= 5\sqrt3 \times \sqrt 3\\ \\ ad= 5\times 3\\ \\ ad=15$

Thus, ad is 15 cm.